(3n+2)(2n-3)=0

Solution for (3n+2)(2n-3)=0 equation:

(3n+2)(2n-3)=0

We multiply parentheses ..

(+6n^2-9n+4n-6)=0

We get rid of parentheses

6n^2-9n+4n-6=0

We add all the numbers together, and all the variables

6n^2-5n-6=0

a = 6; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·6·(-6)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*6}=\frac{-8}{12}
=-2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*6}=\frac{18}{12}
=1+1/2
$